# We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?

We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?

My attempt: For the first draw of 2 dice together, the sample space has 36 possible outcomes and the ones with the same number in both dice are $$(1,1), (2,2)...(6,6)$$ so a total of 6. The probability for one throw is $$P_1 = \frac {6}{36} = \frac {1}{6}$$

Therefore for 2 such throws, the probability to get doubles is $$(\frac {1}{6})^2$$.

The probability to get doubles in 3 out of 6 throws is

$$(\frac {1}{6})^3*(\frac {5}{6})^3$$ ?? Is this correct?

or maybe $$\frac {6^3}{36^6}$$?

Thank you!

• For example, probability of getting one double for two throws is $2(1/6)(5/6)$. Commented Jan 14, 2021 at 9:48

As you observed, the probability of obtaining doubles on any one throw of a pair of dice is $$1/6$$ since $$6$$ of the $$36$$ possible outcomes result in doubles. Since the probability of obtaining doubles is the same for each of the six trials, this is a binomial distribution problem.
The probability of obtaining exactly $$k$$ successes in $$n$$ trials is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $$p^k$$ is the probability of $$k$$ successes, $$(1 - p)^{n - k}$$ is the probability of $$n - k$$ failures, and $$\binom{n}{k}$$ counts the number of ways in which exactly $$k$$ of the $$n$$ trials could result in a success.
Here, we will define a "success" as a double, so $$p = 1/6$$. Since there are six trials, $$n = 6$$. Hence, the probability to get exactly three doubles in six trials is $$\Pr(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$ In your attempt, you did not take into account the number of ways exactly three of the rolls could result in a success.
If you were instead interested in the probability of at least three doubles in six trials, you would add the probabilities of three or more successes. $$\Pr(X \geq 3) = \sum_{k = 3}^{6} \binom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6 - k}$$
• Thank you, I hadn't taken into account the term $\binom{6}{3}$. Commented Jan 14, 2021 at 10:46